Expert: Steve Holleran - 10/8/2007

Question
QUESTION: if an inequality in > or < 0 ... how do you solve it?  
EX:  5X < 0    or      2/3X - 4 > 0

ANSWER: Hi Kelly,

Inequalities are solved pretty much like equations; the only new rule is : if you multiply or divide both sides by a negative number, you have to reverse the order sign.

Here, for 5x < 0, just divide both sides by 5:

         x < 0  This is the solution.


For the second one,   you do two steps:  first add 4, then multiply by the reciprocal of 2/3:

         2/3 * x - 4 > 0

         +4  +4

         2/3 * x > 4

      3/2 * 2/3 * x > 4 * 3/2

         x > 6

I hope this helps you out.

Steve Holleran

---------- FOLLOW-UP ----------

QUESTION: thank you that was helpful, does the same thing apply with inequalities and absolute values? EX: 2!2X+1! + 1 >= 19 ??  note (the exclamation sign represent absolute value

Answer
Hi Kelly,

It's somewhat similar, but with absolute values, you have to deal with positive and negative cases.  Here, try this:

2 |2x + 1| + 1 >= 19

2 |2x + 1| >= 18

  |2x + 1 | >= 9

Now, the quantity in the abs val can be either pos or neg, so here's how to handle the cases:

POS:     2x + 1 >= 9  so 2x >= 8  and x >= 4.

NEG:    -2x -1 >= 9  so -2x >= 10  and x <= -5

(We had to change the order sign because of division by a negative).

Hope this helps out.

Steve